If e^x = 243 and e^y = 32 then e^((3x + 4y)/5) =?

The answer is 432, but I don't understand why.

x = ln 243

y = ln 32

LET Z = e^((3x + 4y)/5)

ln [z ]= (3x+4y)/5

ln z = (1/5)( 3 ln 243 + 4 ln 32)
= ln (243^3/5 *32^4/5)
= ln (27*16)
= ln(432)
if ln z = ln 432
then z = 432

good except for this step:

ln (243^3/5 *32^4/5)
should be
1/5 ln (243^3 * 32^4)

huh?

I agree with Damon's "huh" since

ln (243^3/5 *32^4/5) = 1/5 ln (243^3 * 32^4)

No one is bothered by the fact that 5 does not divide powers of 2 and 3?

just using

ln a^b = b ln a

ln (243^3/5 *32^4/5)
= ln ( (3^5)^(3/5) * (2^5)^(4/5) )
= ln ( 3^3 * 2^4)
= ln (27*16)
= ln(432)

1/5 ln (243^3 * 32^4)
= ln [ (243^3 * 32^4) ^(1/5) ]
= ln (243^(3/5) * 32^(4/5) )
= ....
= ln(432)

Ahh. I see that I was interpreting

243^3/5 as (243^3)/5